Problem: The radius of a circle is increasing at a rate of $3$ centimeters per second. At a certain instant, the radius is $8$ centimeters. What is the rate of change of the area of the circle at that instant (in square centimeters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $9\pi$ (Choice B) B $48\pi$ (Choice C) C $192\pi$ (Choice D) D $64\pi$
Answer: Setting up the math Let... $r(t)$ denote the radius at time $t$, and $A(t)$ denote the circle's area at time $t$. We are given that $r'(t)=3$ and that $r(t_0)=8$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures $A(t)$ and $r(t)$ relate to each other through the formula for the area of a circle: $A(t)=\pi[r(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2\pi r(t)r'(t)$ Using the information to solve Let's plug ${r(t_0)}={8}$ and ${r'(t_0)}={3}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=2\pi{r(t_0)}{r'(t_0)} \\\\ &=2\pi({8})({3}) \\\\ &=48\pi \end{aligned}$ In conclusion, the rate of change of the area of the circle at that instant is $48\pi$ square centimeters per second. Since the rate of change is positive, we know that the area is increasing.